6
$\begingroup$

I assume the twin paradox from special relativity is well known. I wish to focus on the apparent symmetry of the problem: both observer seems to move away from each other, and then come back. Yet, the outcome is asymmetrical. That paradox is resolved because the travelling twin has to turn around at some point. Then he has to change inertial frames, or accelerate, and this means that the situation is not symmetric anymore.

What now if the "turn around" is not caused by an acceleration, but a "gravitational slingshot". Say, the traveling twin's trajectory passes by the graviational field of a couple of stars, freely falling, such that the twins just happen to end up in the same place after a while again. According to general relativity, both twins then remain in an inertial frame. Both see the other initially disappear with slowly ticking clocks, and later reappear from some other direction. All the while, they stayed at rest in their own reference frame. So the other twin must have aged less than themselves. The situation seems completely symmetric again. Of course, since they meet in the same point again, they cannot consider the other one younger from both perspectives.

What is the resolution?

I found this other thread which seems related: Symmetrical twin paradox This refers to a paper that claims that this is due to the fact that the topology of the space induces some preferred reference frame. I don't pretend to understand that paper, but I think it refers to a similar paradox obtained by assuming that the large-scale structure of the universe is spherical/cylindrical, or something like that, allowing one twin to travel "around". Surely it is possible to set up my above scenario with only a few local stars. So it seems strange to me that a local paradox should be solved by relying on a global topologically prefered frame.

$\endgroup$
3
$\begingroup$

"All the while, they stayed at rest in their own reference frame. So the other twin must have aged less than themselves." This is a non sequitur:

There is no reason why if a twin remains in a local inertial frame and encounters twice the the other twin, the latter must be found younger!

It is true in Special Relativity where for a couple of events there is only one geodesic (inertial motion) passing through both, and that curve maximizes the length in the class of timelike curves joining the events. But it could be false in general relativity. One cannot invoke the equivalence principle as it holds in a neighbourhood of a geodesic so it does not imply anything when comparing distant geodesics, even if connected at the ending points.

What you should do is:

(1) Consider the word lines of the twins, namely their stories pictured as timelike curves in spacetime, from the event of their initial separation to the event where they meet again.

(2) Compute the length of these curves using the natural arch length measure in spacetime.

(3) Compare the results.

That length of a world line is a measure of the proper time, i.e. the period of time measured by a clock at rest with the twin (or the physical object describing the considered world line). The longest curve correspond to oldest twin.

If each twin is is freely falling, then each world line is a geodesic and, if the twins meet twice it means that a focal point exists of the class of timelike geodesic emanating from the initial common event.

In general there is no reason why both geodesics must have same length (even if it is possible to prove that there exist a longest geodesic in the set of timelike geodesic connecting the two events).

In case of perfect symmetry (and it is possible for instance in a cylindrical universe) the two twins will have the same age.

$\endgroup$
  • $\begingroup$ Quibble: In a cylindrical universe, the gluing condition imposes a unique global reference frame in which the size of the universe is maximal. There is only perfect symmetry in the case where both observers have equal and opposite relative velocities to this global reference frame. $\endgroup$ – Jerry Schirmer Feb 5 '14 at 17:17
  • $\begingroup$ Yes I agree. Indeed I was just thinking of a 2D cylinder obtained from 2D Minkowsi spacetime where it happens exactly what you remarked. $\endgroup$ – Valter Moretti Feb 5 '14 at 17:20
  • $\begingroup$ I do not think that a nontrivial topology of the spatial section of spacetime is necessary to get focal points for timelike geodesics: think of a pair of rockets orbiting (just gravitation so in inertial motion) around the sun, one clockwise and the other anticlockwise... $\endgroup$ – Valter Moretti Feb 5 '14 at 18:00
  • $\begingroup$ I understand that it is a paradox, not an inconsistency. :) My mental "problem" is with the fact that both observers are in an inertial frame. They therefore see all clocks that follow some other path tick slower (right?). So, if Alice keeps on observing Bob, she should always see his clock progress slower than her own, so when they meet again Bob's clock has fewer ticks. At the same time, when Bob keeps observing Alice, he sees her clock tick slower, with the opposite conclusion. So there must be something in the gravitational field that allows another clock to tick faster. What is that? $\endgroup$ – user40000 Feb 5 '14 at 18:35
  • $\begingroup$ <<both observers are in an inertial frame. They therefore see all clocks that follow some other path tick slower (right?)>> Actually NOT, the point is just this one: It is true in special relativity but in general is false in general relativity. $\endgroup$ – Valter Moretti Feb 5 '14 at 21:51
-1
$\begingroup$

GIVEN: Twin Paradox: One twin travels relativistically, one twin stays home. They reunite. The traveling twin aged much less. The twin who travels through more space accumulates less time; also true for an orbit. Interval sqrt(t^2 - x^2 - y^2 - z^2) between the two events, expressed in inertial coordinate system (t,x,y,z), is conserved. Given the invariant interval, the larger sqrt(x^2 + y^2 + z^2) is the smaller sqrt(t^2) must be.

The ratio by which the two aged when they are again local is identical in all reference frames: ratio = sqrt(t^2 - x^2 - y^2 - z^2)/t (units of c=1).

RESOLUTION: Don't rationalize a boundary condition, remove it. The Twin Paradox without acceleration: Three identical toggle switch mechanical clocks are kits. Three spaceships carry a kit each. Build clocks after the experiment is set up. Toggle touch switches states, on-off. CLOCK 1 is in vacuum free fall, our inertial reference frame, with its toggle exposed, off and zeroed. CLOCK 2's spaceship travels at 0.999c relative to CLOCK 1. It is far to our left. CLOCK 2, off and zeroed, has always been in vacuum free fall. It skims past CLOCK 1, toggles touch, both clocks are on, locally synchronized by touching.

CLOCK 3's spaceship travels at 0.999c relative to CLOCK 1, far far to our right. CLOCK 3 is off and zeroed. CLOCKs 2 and 3, in vacuum free fall, touch toggles. CLOCK 2 is off, CLOCK 3 is on. CLOCK 2's duration is written down. CLOCK 3 reflects CLOCK 2's path to touch toggles with CLOCK 1. Both clocks are off. Write down elapsed times.

Numbers on paper are invariant. Three clocks were passive observers in vacuum free fall (zero acceleration). Compare elapsed times. The Twin (Triplets) Paradox obtains.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.